Three point charges of magnitude 5 mu C,0.16 | vedclass

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(C) Given charges: $q_A = 5 \, \mu C$,$q_B = 0.16 \, \mu C$,$q_C = 0.3 \, \mu C$.Distances: $r_{AB} = 3 \, cm = 3 	imes 10^{-2} \, m$,$r_{AC} = 3 \,

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) Given charges: $q_A = 5 \, \mu C$,$q_B = 0.16 \, \mu C$,$q_C = 0.3 \, \mu C$.Distances: $r_{AB} = 3 \, cm = 3 	imes 10^{-2} \, m$,$r_{AC} = 3 \, cm = 3 	imes 10^{-2} \, m$.Coulomb's Law: $F = \frac{k q_1 q_2}{r^2}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Force on $A$ due to $C$ $(F_1)$: $F_1 = \frac{9 	imes 10^9 	imes (5 	imes 10^{-6}) 	imes (0.3 	imes 10^{-6})}{(3 	imes 10^{-2})^2} = \frac{9 	imes 10^9 	imes 1.5 	imes 10^{-12}}{9 	imes 10^{-4}} = 1.5 	imes 10 = 15 \, N$.Force on $A$ due to $B$ $(F_2)$: $F_2 = \frac{9 	imes 10^9 	imes (5 	imes 10^{-6}) 	imes (0.16 	imes 10^{-6})}{(3 	imes 10^{-2})^2} = \frac{9 	imes 10^9 	imes 0.8 	imes 10^{-12}}{9 	imes 10^{-4}} = 0.8 	imes 10 = 8 \, N$.Since $A$ is the right-angle corner,$F_1$ and $F_2$ are perpendicular.Net force $F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, N$.

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